I need to convert this into a standard conic section equation: $$y=\sqrt{x^2 + r^2 - 2 r x g} + x g$$ $x$ is the independent variable, $r$ is a constant and is always larger than the absolute value of $x$, and $g$ varies between $+1$ and $-1$.
The function's graph looks like a parabola with vertex not at origin.
I would really appreciate a solution or your valuable comments to reduce this equation to something manageable.
$$f(x):=y=\sqrt{x^2 + r^2 - 2 r x g} + x g\tag{*}$$
There is something which is not clear in your definition : "$r > |x|$" ; indeed, I don't see what it means because $x$ is variable...
It isn't a parabola but a branch of hyperbola as you can see it on figure 1.
Fig. 1 : case $r=1/2, \ g=1/4$. Consider only the red branch.
First of all, writing the square root under the form :
$$\sqrt{(x-gr)^2+r^2(1-g^2)}$$
the quantity under the radical sign is always $\ge 0$ due to the fact that $|g| \le 1$. Therefore, function $f$ is defined for any $x \in \mathbb{R}$.
Let us now transform (*) under the form :
$$y-xg=\sqrt{x^2 + r^2 - 2 r x g}$$
which implies by squaring (but isn't equivalent to) :
$$(y-xg)^2=x^2 + r^2 - 2 r x g\tag{1}$$
Indeed, by doing that, a second, "spurious", equation
$$y-xg=\color{red}{-}\sqrt{x^2 + r^2 - 2 r x g}$$
related to the other branch of the hyperbola (in blue on the figure) has "crept in".
From (1),
$$y^2 - 2 xy g + x^2 g^2 = x^2 + r^2 - 2 r x g \ \ \iff \ \ (1-g^2)x^2 -y^2 + 2 xy g - 2 r x g+r^2=0$$
it is easy to get the associated matrix :
$$C=\begin{pmatrix}1-g^2&g&-gr\\g&-1&0\\-gr&0&r^2\end{pmatrix}$$
Let $D$ be the upper left $2 \times 2$ matrix of $C$. As the ratio
$$\det(D)/\det(C)=-1/(r^2(g^2-1))>0$$
we can conclude that we have a hyperbola.